Tìm n thuộc Z :
a, n2 + 5n +9 \(⋮\)n + 5
b, n + 3 \(⋮\)n2 - 7
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\(M=\frac{-2x}{3}+3x\left(\frac{x}{6}-\frac{-2}{9}-\frac{7}{5}\right)-\frac{5x}{2}\left(\frac{x}{5}-\frac{4}{5}\right)\)
\(M=\frac{-2x}{3}+3x\left[\frac{x}{6}-\left(-\frac{2}{9}\right)-\frac{7}{5}\right]-\frac{5x}{4}\left(\frac{x}{5}-\frac{4}{5}\right)\)
\(M=\frac{-2x}{3}+3x\left(\frac{x}{6}+\frac{2}{9}-\frac{7}{5}\right)-\frac{5x}{2}\left(\frac{x}{5}-\frac{4}{5}\right)\)
\(M=-\frac{2x}{3}+3x\left(\frac{x}{6}-\frac{53}{45}\right)-\frac{5x}{2}.\frac{x-4}{5}\)
\(M=-\frac{2x}{3}+3x\left(\frac{x}{6}-\frac{53}{45}\right)-\frac{5x\left(x-4\right)}{10}\)
\(M=-\frac{2x}{3}+3x\left(\frac{x}{6}-\frac{53}{45}\right)-\frac{x\left(x-4\right)}{2}\)
\(M=-\frac{2x}{3}+\frac{x^2}{2}-\frac{53x}{15}-\frac{x\left(x-4\right)}{2}\)
\(M=\left(-\frac{2x}{3}-\frac{53x}{15}\right)+\frac{x^2}{2}-\frac{x\left(x-4\right)}{2}\)
\(M=-\frac{21x}{5}+\frac{x^2}{2}-\frac{x\left(x-4\right)}{2}\)
\(M=\frac{-2.21x+5x^2-5x\left(x-4\right)}{10}\)
\(M=\frac{-42x+5x^2-5x\left(x-4\right)}{10}\)
\(M=\frac{-x\left[42-5x+5\left(x-4\right)\right]}{10}\)
\(M=\frac{-x\left(42-5x+5x-20\right)}{10}\)
\(M=\frac{-x\left(42-20\right)}{10}\)
\(M=\frac{-x.22}{10}\)
\(M=-\frac{22x}{10}\)
\(M=-\frac{11x}{5}\)
\(\left|2x-\frac{1}{2}\right|>\left|-1,5\right|\)
\(\Rightarrow\left|2x-\frac{1}{2}\right|>1,5\)
\(\text{Vì }\left|2x-\frac{1}{2}\right|\ge0\)
\(\Rightarrow2x-\frac{1}{2}>1,5\)
\(\Rightarrow2x-\frac{1}{2}>\frac{15}{10}=\frac{3}{2}\)
\(\Rightarrow2x>2\)
\(\Rightarrow x>1\)
\(\frac{72-x}{3}=\frac{x-18}{5}\)
\(\Rightarrow\frac{\left(72-x\right).5}{15}=\frac{3\left(x-18\right)}{15}\)
\(\Rightarrow\left(72-x\right).5=3\left(x-18\right)\)
\(\Rightarrow360-5x=3x-54\)
\(\Rightarrow-3x-5x=-360-54\)
\(\Rightarrow-8x=-414\)
\(\Rightarrow x=51,75\)
+) Vì \(\frac{72-x}{3}=\frac{x-18}{5}\)
\(\Rightarrow5\left(72-x\right)=3\left(x-18\right)\)
\(5.72-5x=3x-3.18\)
\(360-5x=3x-54\)
\(-5x-3x=-54-360\)
\(-8x=-414\)
\(x=-414:\left(-8\right)\)
\(x=51,75\)
Vậy x = 51,75
Ta có: 2x - 6 = 0 => x = 3
x + 3 = 0 => x = -3
Lập bảng xét dấu:
+) Với x < -3
Ta có: 6 - 2x - x - 3 = 8
=> -3x + 3 = 8
=> -3x = 5
=> x = -5/3 (ko thỏa mãn)
+) Với -3 ≤ x < 3
Ta có: 6 - 2x + x + 3 = 8
=> -x + 9 = 8
=> x = -1
=> x = 1 (thỏa mãn)
+) Với 3 ≤ x
Ta có: 2x - 6 + x + 3 = 8
=> 3x - 3 = 8
=> 3x = 11
=> x = 11/3 (thỏa mãn)
Vậy...
\(\frac{6}{x^2+2}+\frac{12}{x^2+8}=3-\frac{7}{x^2+3}\)
\(\Leftrightarrow6\left(x^2+8\right)\left(x^3+3\right)+12\left(x^2+2\right)\left(x^2+3\right)=3\left(x^2+2\right)\left(x^2+8\right)\left(x^2+3\right)-7\left(x^2+8\right)\left(x^2+2\right)\)
\(\Leftrightarrow18x^4+126x^2+216=3x^6+32x^4+62x^2+32\)
\(\Leftrightarrow18x^2+126x^2+216-3x^6-32x^4-68x^2-32=0\)
\(\Leftrightarrow-14x^4+58x^2+184-3x^6=0\)
\(\Leftrightarrow x=\pm2\)
\(\Rightarrow x=\pm2\)
a) \(4x^2\left(x-2\right)-x+2=0\)
\(\Leftrightarrow4x^2\left(x-2\right)-\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(4x^2-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\4x^2-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=\pm\frac{1}{2}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=\pm\frac{1}{2}\end{cases}}\)
g) \(x^2-7x+10=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\x=2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=5\\x=2\end{cases}}\)
a, Ta có : n2 + 5n + 9 \(⋮\)n + 5
= n (n + 5) + 9 \(⋮\)n + 5
Vì n (n + 5) \(⋮\)n + 5 => 9 \(⋮\)n + 5
=> n + 5 \(\in\)Ư(9) = {\(\pm\)1 ; \(\pm\)3\(\pm\)9 }
Ta lập bảng :
Vậy ...................................................
Thấy đúng thì t.i.c.k đúng cho mik nhé !